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(ai) Increasing the temperature of a liquid decreases the viscosity of the liquid. (aii) Increasing the temperature of a gas increases the viscosity of the gas (b) Factors that determine the magnitude of the moment of a force about a point are: 1. Angle between the distance from the turning point and the line of action of the force. 2. magnitude of the applied force. 3. perpendicular distance from the pivot to the line of action of the applied force. (c) SEE THE DIAGRAM ABOVE. (ii) Taking a moment about the support: clockwise moment = anti-clockwise moment 400 x 10 = m x (\(\frac{L}{2}\) - 10) -- ------- from figure 1 above 4000 = m x (\(\frac{L}{2}\) - 10) - - - -- - - (i) 400 x 6 = m x (\(\frac{L}{2}\) - 15) - - - - -- - - -(ii) divide eqn i by ii \(\frac{4000}{2400}\) = \(\frac{m \times \frac{L}{2} - 10)}{m \times \frac{L}{2} - 15)}\) \(\frac{5}{3}\) = \(\frac{\frac{L}{2} - 10)}{\frac{L}{2} - 15)}\) 3 x (\(\frac{L}{2}\) - 10) = 5 x (\(\frac{L}{2}\) - 15) \(\frac{3L}{2}\) - 30) = \(\frac{5L}{2}\) - 75) \(\frac{5L}{2}\) - \(\frac{3L}{2}\) = 75 - 30 = 45 L = \(\frac{2L}{2}\) = 45 cm From equation (ii) 400 x 6 = m x (\(\frac{L}{2}\) - 15) 2400 = m x (\(\frac{45}{2}\) - 15) 2400 = m x 7.5 m = \(\frac{2400}{7.5}\) = 320 g Therefore, I. m = 320g and II. L = 45 cm (d) Reasons for less pressure at the top of a mountain: The lower pressure at the top of a mountain is due to the reduced number of air molecules in the atmosphere above that elevation, resulting in a lighter air column and fewer molecular collisions, which together create lower atmospheric pressure compared to sea level. (e)(i) Density = \(\rho\) = \(\frac{m}{v}\) m = v x \(\rho\), but v = A x h m = A x h x \(\rho\) = 13.6 x 10\(^3\) x 40 x 20 x 10\(^{-6}\) = 10.88kg (e)(ii) pressure exerted by the surface of the mercury P\(_t\) = P\(_{atm}\) + pressure of mercury P\(_{atm}\) = P\(_t\) - \(\rho\) x h x g = 1.42 x 10\(^5\) - 13.6 x 10\(^3\) x 0.2 x 10 = 1.148 x 10\(^5\)Pa

(a)i The charged particle must be in motion or have velocity and the particle's velocity must be inclined to the magnetic field at an angle \(\theta\) (ii) F = Qv\(\beta\)sin\(\theta\) (iii) for a charged particle moving parallel to the magnetic field, \(\theta\) = 0 and F = Qv\(\beta\)sin\(\theta\) F = Qv\(\beta\)sin 0 = 0 Thus, F = 0 (b)The main parts of an electrical transformer are the coil, Soft iron core (ii) An alternating current in the primary coil produces a changing magnetic field in the primary coil. The changing magnetic field around the core gives rise to a changing magnetic flux linking the secondary coil thereby inducing an alternating e.m.f in the secondary coil as stated in Faraday's law. c(i) Why the speed of the proton remains constant: The speed of the proton remains constant because the magnetic force acts perpendicular to its velocity, providing centripetal acceleration without doing work on it. This means the force changes the direction of the proton's motion but not its speed, resulting in constant kinetic energy and a uniform speed throughout its circular path. (ii) I. radius of curved path qv\(\beta\) = \(\frac{mv^2}{r}\) r = \(\frac{mv}{q\beta}\) r = \(\frac{1.7 \times 10^{-27}}{1.6 \times 10^{-19}}\) = 0.24 m. II. time taken by the proton to move through the magnetic field, t = \(\frac{\pi r}{2v}\) = \(\frac{3.14 \times 0.24}{2 \times 4.5 \times 10^6}\) = 8.37 x 10\(^{-8}\)s

(v) M(g) W\(_A\)(N) W\(_W\)(N) U\(_1\) = W\(_A\) - W\(_W\). W\(_L\)(N) U\(_2\) = W\(_A\) - W\(_L\) 20.0 0.20 0.16 0.04 0.18 0.02 40.0 0.40 0.33 0.07 0.36 0.04 60.0 0.60 0.49 0.11 0.54 0.06 80.0 0.80 0.66 0.14 0.72 0.08 100.0 1.00 0.82 0.18 0.90 0.10 (vi) SEE THE GRAPH ABOVE (vii) s = \(\frac{\Delta U_1}{\Delta U_1}\) = \(\frac{0.162 - 0.2}{0.09 - 0.067}\) = \(\frac{0.042}{0.023}\) = 1.826 (viii) K = \(\frac{1}{\text{s}}\) = \(\frac{1}{1.826}\) = 0.5476 (ix)(I) I avoided errors due to parallax when reading the spring balance. (II) I cleared the mass thoroughly before immersing it in liquid L b(i) Archimedes's principle states that when a body is completely or partially immersed in a fluid, it experiences an upthrust which is equal to the weight of fluid displaced. (ii) Density has a unit whereas relative density has none. (II) The value of density is different in different systems of measurement, whereas relative density is the same in all systems of measurement.

(vii) L(cm) L\(_x\)(cm) L\(_y\)(cm) R\(_1\) = (\(\frac{L_y}{L_x})R_x\)\(\Omega\) L\(_{x2}\) L\(_{y2}\) R\(_2\) = \(\frac{L_{y2}}{L_{x2}}R_x\)\(\Omega\) 100.0 5.2 94.8 18.2 21.5 78.5 3.6 90.0 6.0 94.0 15.7 22.5 77.5 3.4 80.0 6.8 93.2 13.7 23.4 76.6 3.3 70.0 7.6 92.4 12.2 24.5 75.5 3.1 60.0 8.4 91.6 10.9 25.6 74.4 2.9 (viii) SEE THE GRAPH ABOVE (ix) Slope = \(\frac{\Delta R_2}{\Delta R_1}\) = \(\frac{3.5 - 3.3}{17.0 - 14.6}\) = \(\frac{0.2}{2.4}\) = 0.083 (x) K = \(\sqrt{0.083}\) = 0.2881 (xi) I Tight connection/clean terminal was ensured II. Key removed when readings were not being taken. b(i) I. A potentiometer can be used for measuring the internal resistance of a cell and also to compare the e.m.f. of two cells. II. A potentiometer gives the exact value of potential difference across any two points in a circuit. (ii) It refers to the opposition to the flow of current offered by the cells and batteries themselves resulting in the generation of heat.

Dry Leclanché: This is a common non-rechargeable cell.

(a)i c = 50.0cm (ii) M\(_R\) = 90.0 g (vii) d = 0.8m (ix) S/N h(cm) t(s) T= \(\frac{t}{20}\)s T\(^2\)s\(^2\) 1 40.0 26.12 1.306 1.7056 2 50.0 27.80 1.390 1.9321 3 60.0 30.68 1.534 2.3532 4 70.0 34.60 1.730 2.9929 5 80.0 35.80 1.790 3.2041 (x) SEE THE GRAPH ABOVE (xi) Slope, s = \(\frac{\DeltaT^2}{\Delta h}\) = \(\frac{3.0 - 1.75}{75 - 46}\) = \(\frac{1.25}{29}\) = 0.431s\(^2\)cm\(^{-1}\) (xii) K = \(\frac{\text{S}}{\text{Q}}\) where, Q = \(\frac{2}{25d^2}\) = \(\frac{2}{25 \times (0.8)^2}\) = 0.125 K = \(\frac{\text{S}}{\text{Q}}\) = \(\frac{0.431}{0.125}\) = 0.3448 b(i) A couple, as related to oscillatory motion, refers to a pair of forces that are equal in magnitude but opposite in direction, and act on opposite sides of a pivot or fulcrum, causing rotation or torsion. b(ii) I. Wrenches: When using a wrench to tighten or loosen a bolt, the force applied creates a couple that produces rotational motion. II. Steering Wheels: Turning the steering wheel of a car applies a couple that rotates the front wheels, allowing for directional change. III. Door Handles: When pushing or pulling on one side of a door handle, a couple is created that rotates the door about its hinges. IV. Bicycle Pedals: When pedaling a bicycle, the force applied to the pedals generates a couple that turns the crankshaft, propelling the bike forward.

The question can be explained by Graham's law of diffusion. According to this law, under the same conditions of temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. \(\frac{R_x}{R_y}\) = \(\sqrt{\frac{M_y}{M_x}}\) From the question, Gas X diffuses twice as fast as gas Y i.e x = 2y This implies x =2 while y = 1 From \(\frac{R_x}{R_y}\) = \(\sqrt{\frac{M_y}{M_x}}\) \(\frac{2}{1}\) = \(\sqrt{\frac{M_y}{M_x}}\) \(\frac{4}{1}\) = \(\frac{M_y}{M_x}\) M\(_y\) = 4 M\(_x\) ⇒ Gas Y is 4 times as heavy as Gas X. - Option B

To find the number of molecules, we'll use the formula: Number of moles = \(\frac{Number of molecules}{Avogadro number}\) But We were only given the mass of Helium = 11.5g , knowing the atomic mass or molar mass of He = 4 g/mol Number of moles = \(\frac{mass}{molar mass}\) = \(\frac{11.5}{4}\) = 2.875 moles Number of molecules = Number of moles x Avogadro number = 2.875 x 6.02 x 10\(^{23}\) = 17.3075 x 10\(^{23}\) = 1.73 x 10\(^{24}\)

Alnico is ferromagnetic, meaning it can be used to make permanent magnets. Alnico is an iron alloy with the acronym "al-ni-co" that's made up of aluminum (Al), nickel (Ni), and cobalt (Co). Option B

Gold and silver are noble metals, because they are resistant to corrosion and oxidation, even at high temperatures. Noble metals are a group of metals that are less reactive than other metals and are not easily corroded or oxidized.

Aluminum is extracted from bauxite by electrolysis. The extraction proceeds in two stages; 1. Purification of the Bauxite: The impure bauxite is heated with sodium hydroxide solution to form soluble sodium tetrahydroxy aluminate (iii). The impurities in the ore which are iron (iii) oxide and trioxosilicate (iv) compounds are not soluble in the alkali. They are therefore filtered off as a sludge. Aluminum hydroxide crystals is then added to filtrate, NaAl(OH)\(_4\) solution to induce the precipitation of Aluminum hydroxide. 2. The electrolysis of the pure alumina

In contact with water, calcium carbide instantly decomposes hydrolytically, yielding acetylene gas and calcium hydroxide. Calcium ethynide is also known as calcium carbide. CaC\(_2\) + 2H\(_2\)O → Ca(OH)\(_2\) + C\(_2\)H\(_2\)

Bitumen is the petroleum product which is used for surfacing roads. It is also known as asphalt. It is used as an alternative for coal-tar.​

The petroleum fractions that distill at 200–250^\circC are naphtha and kerosene, Bitumen: has a boiling point value of 525\(^0\)C Gasoline: Has a boiling range of 40–200^\circC Diesel: Has a boiling range of 250–350^\circC Fuel oil, paraffin, lubricants: Have a boiling range of 300–370^\circC Residue, crude oil: Has a boiling point of more than 370^\circC

Coal tar is a known human carcinogen. It contains potent carcinogens like benzo(a)pyrene, benz(a)anthracene, and dibenz(a,h)anthracene. Fractional distillation of coal tar is a process that produces aromatic hydrocarbons like benzene, creosotes, naphthalene, anthracene, carbolic and cresylic acids, pyridine, and pitch. Benzene is used in the pharmaceutical industry as a solvent and in the synthesis of drugs such as cyclohexane, phenol etc.Thus the correct answer is option B- Benzene